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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation of the family of circles touching the \(y\) - axis at origin.

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  • It is given that the circle touches the y-axis at origin
  • Hence the equation of the circle is $(x-a)^2 + y^2 = a^2$
Step 1:
$(x-a)^2 + y^2 = a^2$
On simplifying we get,
$x^2 - 2ax +a^2+ y^2= a^2$
$x^2 + y^2 = 2ax$------(1)
Step 2:
Differentiating on both sides we get,
$2x + 2yy' = 2a$
dividing throughout by 2 we get
$x + yy' = a$-----(2)
Step 3:
Substituting equ(2) in equ(1) we get
$x^2 + y^2 = 2(x+yy')x$
On simplifying and rearranging we get
$x^2 + 2xyy' = y^2$
This is the required equation.
answered Aug 16, 2013 by sreemathi.v

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